洛谷P3586 [POI2015]LOG(贪心 权值线段树)

题意

题目链接

sol

显然整个序列的形态对询问没什么影响

设权值\(>=s\)的有\(k\)个。

我们可以让这些数每次都被选择

那么剩下的数，假设值为\(a_i\)次，则可以\(a_i\)次被选择

一个显然的思路是每次选最大的c个

那么只需要判断\(\sum a_i >=(c - k)*s\)即可

权值线段树维护一下

#include<bits/stdc++.h> 
#define pair pair<int, int>
#define mp(x, y) make_pair(x, y)
#define fi first
#define se second
#define ll long long 
#define fin(x) {freopen(#x".in","r",stdin);}
#define fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int maxn = 1e6 + 10, mod = 1e9 + 7, inf = 1e9 + 10;
const double eps = 1e-9;
template <typename a, typename b> inline bool chmin(a &a, b b){if(a > b) {a = b; return 1;} return 0;}
template <typename a, typename b> inline bool chmax(a &a, b b){if(a < b) {a = b; return 1;} return 0;}
template <typename a, typename b> inline ll add(a x, b y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename a, typename b> inline void add2(a &x, b y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename a, typename b> inline ll mul(a x, b y) {return 1ll * x * y % mod;}
template <typename a, typename b> inline void mul2(a &x, b y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename a> inline void debug(a a){cout << a << '\n';}
template <typename a> inline ll sqr(a x){return 1ll * x * x;}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int a[maxn], n, m;
const int ss = maxn * 10 + 10, mx = 1e9 + 10;
int ls[ss], rs[ss], num[ss], tot, root;
ll sum[ss];
void update(int k) {
    sum[k] = sum[ls[k]] + sum[rs[k]];
    num[k] = num[ls[k]] + num[rs[k]];
}
void modify(int &k, int l, int r, int p, int v) {
    if(!k) k = ++tot;
    if(l == r) {
        num[k] += (v < 0 ? -1 : 1);
        sum[k] += v;
        return ;
    }
    int mid = l + r >> 1;
    if(p <= mid) modify(ls[k], l, mid, p, v);
    else modify(rs[k], mid + 1, r, p, v);
    update(k);
}
int querynum(int k, int l, int r, int ql, int qr) {
    if(!k) return 0;
    if(ql <= l && r <= qr) 
        return num[k];
    int mid = l + r >> 1;
    if(ql > mid) return querynum(rs[k], mid + 1, r, ql, qr);
    else if(qr <= mid) return querynum(ls[k], l, mid, ql, qr);
    else return querynum(ls[k], l, mid, ql, qr) + querynum(rs[k], mid + 1, r, ql, qr);
}
ll querysum(int k, int l, int r, int ql, int qr) {
    if(!k) return 0;
    if(ql <= l && r <= qr) return sum[k];
    int mid = l + r >> 1;
    if(ql > mid) return querysum(rs[k], mid + 1, r, ql, qr);
    else if(qr <= mid) return querysum(ls[k], l, mid, ql, qr);
    else return querysum(ls[k], l, mid, ql, qr) + querysum(rs[k], mid + 1, r, ql, qr);
}
signed main() {
    n = read(); m = read();
    while(m--) {
        char s[3]; scanf("%s", s);
        int x = read(), y = read();
        if(s[0] == 'u') {//a[x] = y
            if(a[x]) modify(root, 1, mx, a[x], -a[x]);
            a[x] = y;
            if(y) modify(root, 1, mx, y, y);
        } else {//choose x = c   turn y = s
            int k = querynum(1, 1, mx, y, mx);
            ll sum = querysum(1, 1, mx, 1, y - 1);
            puts((sum >= 1ll * (x - k) * y) ? "tak" : "nie");
        }
    }
    
    
    return 0;
}
/*
7
-1 160 -2000
14 82 61 85 41 10 34
*/